We sketch the proof of the easy half of the theorem.

This proves one half of (2); the other half is a matter of direct computation.

the left half of the interval

the upper half of the unit disc

Thus $F$ is greater by a half.


$F$ is half the sum of the positive roots. [Or: half of the sum]

Half of the sets of $R$ miss $i$ and half the remaining miss $j$.

Then $J$ contains an interval of half its length in which $f$ is positive.

On average, about half the list will be tested.

Now $F$ is half as long as $G$. [Or: as $G$ is]

We divide $N$ in half.

The length of $F$ is thus reduced by half.

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